# integral to find the volume of the solid

Use a triple integral to find the volume of the solid bounded by the parabolic cylinder and the planes and

The range of z is obviously from z = 0 to z = 6.

The range of y is obviously from y = 0 to y = 5, with the lower bound being 0 because $2x^2$ us always greater than 0.

The range of x is a little trickier but can be easily determined by solving $y = 2x^2$ and determining the upper and lower expressions for the value of x.

$$V = \int_0^6\int_0^5\int_{-\sqrt{\frac{y}{2}}}^\sqrt{\frac{y}{2}} dxdydz\, \\ V = \int_0^6\int_0^5\ \sqrt{\frac{y}{2}} - (- \sqrt{\frac{y}{2}}) dydz\,\\ V = \int_0^6\int_0^5\ \sqrt{2}\sqrt{y} dydz\,\\ V = \int_0^6 \sqrt{2}\frac{2}{3}y^\frac{3}{2} \left.\right\vert_{0}^{5}dz\,\\ V = \int_0^6 \sqrt{2}\frac{2}{3}\sqrt{5^3}dz\,\\ V = 6\sqrt{2}\frac{2}{3}\sqrt{5^3}\,\\ V = 20\sqrt{10} \\$$

Note that the value being integrated (after the bounds have been determined) is 1. This could change if you're integrating over some density function where the weighting of every point with the object will vary.

• well, now after finding the limits what do we integrate??? Mar 13, 2014 at 3:47
• @user131040 You integrate 1. dxdydz = 1dxdydz Mar 13, 2014 at 3:52
• Oh, sorry if that wasn't clear. I'm in the process of adding my work but @zerosofthezeta is correct. Mar 13, 2014 at 3:53

The integral set up for the volume V is:

$$\int_ 0 ^5\int_{-(y/2)^{.5} } ^{(y/2)^{(.5)}}\int_ 0 ^ 6dzdxdy$$. You can

integrate with respect to z first, then x , and then y.

• With a reputation of 2k+, I wonder why you did not use LaTex? Was it on purpose? Mar 13, 2014 at 3:38
• @zerosofthezeta: brutally honestly, I read a few tutorial pages but need a tutor to sit down with me to walk me through it. Mar 13, 2014 at 3:40
• I got 2sqrt10 for the final answer which is wrong! Mar 13, 2014 at 3:40
• @user131040: will fix it Mar 13, 2014 at 3:46